If x = 2, y = 3 and z = 5, calculate the following: 5x - z

Mathematics

1 answer:

nikdorinn [45]2 years ago

5 0

5*2 - 5
5*2=10
10-5=5

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How many solutions does the system of equations have?

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=============
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7y - 49x = -2
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3 years ago

Solve the equation in the interval [0,2π]. If there is more than one solution write them separated by commas.

Sedaia [141]

$\large\begin{array}{l} \textsf{Solve the equation for x:}\\\\ 
\mathsf{(tan\,x)^2+2\,tan\,x-4.76=0}\\\\\\ \textsf{Substitute}\\\\ 
\mathsf{tan\,x=t\qquad(t\in \mathbb{R})}\\\\\\ \textsf{so the equation 
becomes}\\\\ \mathsf{t^2+2t-4.76=0}\quad\Rightarrow\quad\begin{cases} 
\mathsf{a=1}\\\mathsf{b=2}\\\mathsf{c=-4.76} \end{cases} 
\end{array}$

$\large\begin{array}{l} \textsf{Using 
the quadratic formula:}\\\\ \mathsf{\Delta=b^2-4ac}\\\\ 
\mathsf{\Delta=2^2-4\cdot 1\cdot (-4.76)}\\\\ 
\mathsf{\Delta=4+19.04}\\\\ \mathsf{\Delta=23.04}\\\\ 
\mathsf{\Delta=\dfrac{2\,304}{100}}\\\\ 
\mathsf{\Delta=\dfrac{\diagup\!\!\!\! 4\cdot 576}{\diagup\!\!\!\! 4\cdot
 25}}\\\\ \mathsf{\Delta=\dfrac{24^2}{5^2}} \end{array}$

$\large\begin{array}{l}
 \mathsf{\Delta=\left(\dfrac{24}{5}\right)^{\!2}}\\\\ 
\mathsf{\Delta=(4.8)^2}\\\\\\ 
\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\ 
\mathsf{t=\dfrac{-2\pm\sqrt{(4.8)^2}}{2\cdot 1}}\\\\ 
\mathsf{t=\dfrac{-2\pm 4.8}{2}}\\\\ \mathsf{t=\dfrac{\diagup\!\!\!\! 
2\cdot (-1\pm 2.4)}{\diagup\!\!\!\! 2}}\\\\\mathsf{t=-1\pm 2.4} 
\end{array}$

$\large\begin{array}{l} \begin{array}{rcl} 
\mathsf{t=-1-2.4}&~\textsf{ or }~&\mathsf{t=-1+2.4}\\\\ 
\mathsf{t=-3.4}&~\textsf{ or }~&\mathsf{t=1.4} \end{array} 
\end{array}$

$\large\begin{array}{l} \textsf{Both 
are valid values for t. Substitute back for }\mathsf{t=tan\,x:}\\\\ 
\begin{array}{rcl} \mathsf{tan\,x=-3.4}&~\textsf{ or 
}~&\mathsf{tan\,x=1.4} \end{array}\\\\\\ \textsf{Take the inverse 
tangent function:}\\\\ \begin{array}{rcl} 
\mathsf{x=tan^{-1}(-3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi}\\\\ 
\mathsf{x=-tan^{-1}(3.4)+k\cdot \pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+k\cdot \pi} \end{array}\\\\\\ 
\textsf{where k in an integer.} \end{array}$

__________

$\large\begin{array}{l}
 \textsf{Now, restrict x values to the interval 
}\mathsf{[0,\,2\pi]:}\\\\ \bullet~~\textsf{For }\mathsf{k=0:}\\\\ 
\begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=1:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+\pi} 
\end{array}}\textsf{ is in the 2}^{\mathsf{nd}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 1.86~rad~~(106.39^\circ)}\\\\\\ 
\boxed{\begin{array}{c}\mathsf{x=tan^{-1}(1.4)+\pi} \end{array}}\textsf{
 is in the 3}^{\mathsf{rd}}\textsf{ quadrant.}\\\\ \mathsf{x\approx 
4.09~rad~~(234.46^\circ)}\\\\\\ \end{array}$

$\large\begin{array}{l}
 \bullet~~\textsf{For }\mathsf{k=2:}\\\\ \begin{array}{rcl} 
\mathsf{x=-tan^{-1}(3.4)+2\pi}&~\textsf{ or 
}~&\mathsf{x=tan^{-1}(1.4)+2\pi>2\pi~~\textsf{(discard)}} 
\end{array}\\\\\\ \boxed{\begin{array}{c}\mathsf{x=-tan^{-1}(3.4)+2\pi} 
\end{array}}\textsf{ is in the 4}^{\mathsf{th}}\textsf{ quadrant.}\\\\ 
\mathsf{x\approx 5.00~rad~~(286.39^\circ)} \end{array}$

$\large\begin{array}{l}
 \textsf{Solution set:}\\\\ 
\mathsf{S=\left\{tan^{-1}(1.4);\,-tan^{-1}(3.4)+\pi;\,tan^{-1}(1.4)+\pi;\,-tan^{-1}(3.4)+2\pi\right\}}
 \end{array}$

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$\large\textsf{I hope it helps.}$

Tags: <em>trigonometric trig quadratic equation tangent tan solve inverse symmetry parity odd function</em>

6 0

3 years ago

A region R is enclosed by the coordinate graph of y=k(x-5)^2. When this region is revolved around the x-axis, the solid has a vo

alexandr1967 [171]

K would equal 3.

Your welcome.

4 0

3 years ago